Question

1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated in steam gave 2.50 g of its oxide. Show that these results illustrate the law of constant proportion.​

Answer 1

• 1.50 g of the same metal heated in steam gave 2.50 g of its oxide.
• Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen are fixed.
• Hence, the results follows the law of constant proportion.

Answer 2

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▌Given ▌

• 1.80 g of a certain metal is burnt in oxygen, which gave 3.0 g of its oxide.
• 1.50 g of the same metal is heated  in steam, which gave 2.50 g of its oxide.

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▌To Prove ▌

These results illustrate the law of constant proportion.​

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▌Solution ▌

In the first sample of the oxide,

• Weight of metal = 1.80 g
• Weight of Oxygen = (3.0 - 1.80) = 1.20 g

$\sf{\dfrac{Wt.\ of\ metal}{Wt.\ of\ Oxygen}=\dfrac{1.80g}{1.20g} =1.5 }$

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In the second sample of the oxide,

• Weight of metal = 1.50 g
• Weight of Oxygen = (2.50 - 1.50) = 1 g

$\sf{\dfrac{Wt.\ of\ metal}{Wt.\ of\ Oxygen}=\dfrac{1.50g}{1g} =1.5 }$

Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen are fixed.

Hence, the results follows the law of constant proportion.